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in Mathematics by (106k points)

If D is the mid-point of the hypotenuse AC of a right triangle ABC, prove that BD = 1/2 AC.

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Let ∆ABC is a right triangle such that ∠B = 900 and D is mid point of AC then we have to prove that BD = 1/2 AC we produce BD to E such that BD = AC and EC.  

Now is ∆ADB and ∆CDE we have 

AD = DC [Given] 

BD = DE [By construction] 

And, ∠ADB = ∠CDE [Vertically opposite angles] 

∴ By SAS criterion of congruence we have 

∆ADB ≅ ∆CDE 

⇒ EC = AB and ∠CED = ∠ABD ....(i) [By cpctc] 

But ∠CED & ∠ABD are alternate interior angles 

∴ CE ║ AB ⇒ ∠ABC + ∠ECB = 1800 [Consecutive interior angles] ⇒ 90 + ∠ECB = 1800 

⇒ ∠ECB = 900 

Now, In ∆ABC & ∆ECB we have 

AB = EC [By (i)] 

BC = BC [Common] 

And, ∠ABC = ∠ECB = 900 

∴ BY SAS criterion of congruence 

∆ABC ≅ ∆ECB 

⇒ AC = EB [By cpctc] 

⇒ 1/2 AC = 1/2 EB 

⇒ BD = 1/2 AC 

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